Problem: The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered. 

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 103 ≤ x, y ≤ 103), and r (1 ≤ r ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line. 

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10
60 0 10
30 30 10


Output

30.00000 0.00000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class _T>
inline _T sqr(const _T &x) {
  return x * x;
}
template <class _T>
inline string tostr(const _T &a) {
  ostringstream os("");
  os << a;
  return os.str();
}
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-11;
struct tp {
  long double x, y;
};
struct tl {
  long double a, b, c;
};
inline long double dist(tp p1, tp p2) {
  return sqrt(sqr(p1.x - p2.x) + sqr(p1.y - p2.y));
}
inline long double dist2(tp p1, tp p2) {
  return (sqr(p1.x - p2.x) + sqr(p1.y - p2.y));
}
tl findline(tp p1, tp p2) {
  tl l;
  l.a = p2.y - p1.y;
  l.b = p1.x - p2.x;
  l.c = -l.a * p1.x - l.b * p1.y;
  return l;
}
tl findperp(tl l, tp p) {
  tl l1;
  l1.a = l.b;
  l1.b = -l.a;
  l1.c = -l1.a * p.x - l1.b * p.y;
  return l1;
}
long double distline(tl l, tp p) {
  return (l.a * p.x + l.b * p.y + l.c) / sqrt(l.a * l.a + l.b * l.b);
}
void norm(tl &l) {
  long double d = sqrt(l.a * l.a + l.b * l.b);
  l.a /= d;
  l.b /= d;
  l.c /= d;
}
bool onseg(tp p, tp p1, tp p2) {
  tl l = findline(p1, p2);
  if (fabs(distline(l, p)) > EPS) return false;
  return fabs(dist(p1, p2) - dist(p, p1) - dist(p, p2)) < EPS;
}
long double vp(tp p, tp p1, tp p2) {
  p1.x -= p.x;
  p1.y -= p.y;
  p2.x -= p.x;
  p2.y -= p.y;
  return p1.x * p2.y - p1.y * p2.x;
}
long double sp(tp p, tp p1, tp p2) {
  p1.x -= p.x;
  p1.y -= p.y;
  p2.x -= p.x;
  p2.y -= p.y;
  return p1.x * p2.x + p1.y * p2.y;
}
bool findpoint(tl l1, tl l2, tp &p) {
  long double d = l1.a * l2.b - l1.b * l2.a;
  if (fabs(d) < EPS) return false;
  p.x = (l1.b * l2.c - l1.c * l2.b) / d;
  p.y = (l1.c * l2.a - l1.a * l2.c) / d;
  return true;
}
int findcircline(tp c, long double r, tl l, tp *ps) {
  long double h = fabs(distline(l, c));
  if (h > r - EPS) return 0;
  tl l1 = findperp(l, c);
  tp p;
  long double d = sqrt(l.a * l.a + l.b * l.b);
  h = sqrt(fabs(r * r - h * h));
  findpoint(l, l1, p);
  ps[0].x = p.x + l1.a * h / d;
  ps[0].y = p.y + l1.b * h / d;
  ps[1].x = p.x - l1.a * h / d;
  ps[1].y = p.y - l1.b * h / d;
  return (dist(ps[0], ps[1]) > EPS) + 1;
}
int findcirccirc(tp c1, long double r1, tp c2, long double r2, tp *ps) {
  tl l;
  l.a = 2.0 * (c2.x - c1.x);
  l.b = 2.0 * (c2.y - c1.y);
  l.c =
      c1.x * c1.x + c1.y * c1.y - c2.x * c2.x - c2.y * c2.y + r2 * r2 - r1 * r1;
  return findcircline(c1, r1, l, ps);
}
tp shift(tp p1, tp p2, long double d) {
  tp p;
  long double h = dist(p1, p2);
  p.x = p1.x + (p2.x - p1.x) / h * d;
  p.y = p1.y + (p2.y - p1.y) / h * d;
  return p;
}
istream &operator>>(istream &is, tp &p) {
  is >> p.x >> p.y;
  return is;
}
int main() {
  cout << setiosflags(ios::fixed) << setprecision(5);
  tp a[3];
  long double r[3];
  for (int i = 0; i < (int)(3); i++) {
    cin >> a[i] >> r[i];
  }
  for (int i = 0; i < (int)(3); i++) {
    for (int j = 0; j < (int)(i); j++) {
      if (r[j] > r[i]) {
        swap(r[i], r[j]);
        swap(a[i], a[j]);
      }
    }
  }
  if (fabs(r[0] - r[2]) < EPS) {
    tl l[2];
    tl lp[2];
    for (int i = 0; i < (int)(2); i++) {
      l[i] = findline(a[i], a[i + 1]);
      tp p;
      p.x = (a[i].x + a[i + 1].x) * 0.5;
      p.y = (a[i].y + a[i + 1].y) * 0.5;
      lp[i] = findperp(l[i], p);
    }
    tp pp;
    findpoint(lp[0], lp[1], pp);
    cout << pp.x << " " << pp.y << endl;
    return 0;
  }
  if (fabs(r[1] - r[2]) < EPS) {
    swap(r[0], r[2]);
    swap(a[0], a[2]);
  }
  tp pc[2];
  long double rc[2];
  for (int i = 0; i < (int)(2); i++) {
    long double mi = r[i];
    long double dd = dist(a[i], a[2]);
    long double ma = dd - r[2];
    for (int tt = 0; tt < (int)(239); tt++) {
      long double q = (mi + ma) * 0.5;
      long double a0 = asin(r[i] / q);
      long double a1 = asin(r[2] / (dd - q));
      if (a0 > a1)
        mi = q;
      else
        ma = q;
    }
    tp pp1 = shift(a[i], a[2], mi);
    tp p1 = a[i];
    long double r1 = r[i];
    tp p2 = a[2];
    long double r2 = r[2];
    if (r1 < r2) {
      swap(r1, r2);
      swap(p1, p2);
    }
    mi = dd + r2;
    ma = dd + r2 + 1e+8;
    for (int tt = 0; tt < (int)(239); tt++) {
      long double q = (mi + ma) * 0.5;
      long double a0 = asin(r1 / q);
      long double a1 = asin(r2 / (q - dd));
      if (a0 > a1)
        ma = q;
      else
        mi = q;
    }
    tp pp2 = shift(p1, p2, mi);
    pc[i].x = (pp1.x + pp2.x) * 0.5;
    pc[i].y = (pp1.y + pp2.y) * 0.5;
    rc[i] = dist(pc[i], pp1);
  }
  tp ps[2];
  int cnt = findcirccirc(pc[0], rc[0], pc[1], rc[1], ps);
  long double ans = 1e+20;
  tp pa;
  pa.x = -1.0;
  pa.y = -1.0;
  for (int i = 0; i < (int)(cnt); i++) {
    long double d = dist(ps[i], a[0]);
    if (d < ans) {
      ans = d;
      pa = ps[i];
    }
  }
  if (ans < 1e+19) {
    cout << pa.x << " " << pa.y << endl;
  }
  return 0;
}